Question

two identical masses are as shown in diagram . one is thrown upwards with velocity 20m/s and another is just dropped simultaneously. if the collision between them is elastic , find the time interval between their striking with ground​

Answer 1

The time interval between their striking with ground​ is 2 seconds

Explanation:

Note: Here height h = 20 m

Let the masses meet each other after time t

Velocity of mass thrown upwards after time t

Using the first equation of motion

$v=u-gt$

Here u = 20 m/s

Thus

$v=20-gt$

Distance travelled in time t

$s=ut-\frac{1}{2}gt^2$

$\implies s=20t-\frac{1}{2}gt^2$

Velocity of the mass dropped below after time t

$v'=0+gt$

$\implies v'=gt$

Distance travelled in time t

$s'=0\times t+\frac{1}{2}gt^2$

$s'=\frac{1}{2}gt^2$

$s+s'=h=20$ m

$\implies 20t-\frac{1}{2}gt^2+\frac{1}{2}gt^2=h$

$\implies t=\frac{h}{20}=\frac{20}{20}=1$ second

Therefore

$s=20\times 1-\frac{1}{2}\times 10\times 1^2$

$\implies s=20-5=15$ m

After time t the masses collide, as the collision is elastic and the masses are identical, the velocities of the masses will get exchange

Thus the velocity of the mass thrown upwards will be $v'=gt=10$ m/s in downwards direction and the displacement will be s

Let the time taken by this mass is T

And the velocity of mass dropped downwards will be $v=20-gt=20-10=10$m/s in upwards direction, the displacement of this mass is again s

If the time taken is T'

Then,

$s=10T+\frac{1}{2}gT^2$

And

$s=-10T'+\frac{1}{2}gT'^2$

From above two equations

$10T+\frac{1}{2}gT^2=-10T'+\frac{1}{2}gT'^2$

$\implies 10(T+T')=\frac{1}{2}\times 10(T'^2-T^2)$

$\implies 2(T+T')=(T'-T)(T'+T)$

$\implies T'-T=2$

Therefore, time interval = 2 seconds

Hope this answer is helpful.

Know More:

Q: Two balls are thrown simultaneously,A vertically upwards with a speed of 20m/s from the ground and B vertically downwards from a height of 40m with the same speed and along th same line of motion at what paints do the two balls collide?

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